# Successive numbers starting from 1 are written in a rectangular grid, starting in the top left corner and snaking down diagonally to the right as shown. In which row and column does 2008 occur?

## $\textcolor{w h i t e}{0} 1 \textcolor{w h i t e}{00} 2 \textcolor{w h i t e}{00} 6 \textcolor{w h i t e}{00} 7 \textcolor{w h i t e}{0} 15 \textcolor{w h i t e}{0} 16$ $\textcolor{w h i t e}{0} 3 \textcolor{w h i t e}{00} 5 \textcolor{w h i t e}{00} 8 \textcolor{w h i t e}{0} 14 \textcolor{w h i t e}{0} \text{etc}$ $\textcolor{w h i t e}{0} 4 \textcolor{w h i t e}{00} 9 \textcolor{w h i t e}{0} 13$ $10 \textcolor{w h i t e}{0} 12$ $11$

Feb 6, 2017

The $9$th row, $55$th column.

#### Explanation:

The last number added to the $n$th (reverse) diagonal is the $n$th triangular number ${T}_{n} = \frac{1}{2} n \left(n + 1\right)$.

What is the smallest triangular number greater than or equal to $2008$?

Given:

$\frac{1}{2} n \left(n + 1\right) = {T}_{n} \ge 2008$

Multiply both ends by $2$ to get:

$n \left(n + 1\right) \ge 4016$

So the value of $n$ we are looking for is roughly $\sqrt{4016} \approx 63$ and we find:

${T}_{63} = \frac{1}{2} \cdot 63 \cdot 64 = 2016$

${T}_{62} = \frac{1}{2} \cdot 62 \cdot 63 = 1953$

So $2008$ lies on the reverse diagonal which has ${T}_{63} = 2016$ at one end.

Note that due to the boustrophedonic (like an ox ploughing a field) way in which the numbers are written, ${T}_{63}$ will be written on the top row as the $63$rd term of that row.

$2008$ will occur $8$ rows below and to the left of it, i.e. on the $9$th row in the $55$th column.